Which of the following is true when $x=\mathrm{SinA}+\mathrm{CosA}$; $y=\mathrm{SecA}+\mathrm{CosecA}$?

The correct option is:
$y+2x=x^{2}y$

Let us establish the relationship between $x$ and $y$ step-by-step.

We are given:
$x=\sin A+\cos A$

Squaring both sides of this equation:
$x^2={\left(\sin A+\cos A\right)}^2$
$x^2={\sin }^{2}A+{\cos }^{2}A+2\sin A\cos A$

Using the standard trigonometric identity ${\sin }^{2}A+{\cos }^{2}A=1$, we can rewrite this as:
$x^2=1+2\sin A\cos A$

Rearranging the equation to solve for the product term:
$2\sin A\cos A=x^2-1$
$\sin A\cos A=\frac{x^2-1}{2}$ --- (Equation 1)

Next, we are given:
$y=\sec A+\mathrm{cosec}A$

Expressing secant and cosecant in terms of sine and cosine:
$y=\frac{1}{\cos A}+\frac{1}{\sin A}$

Taking a common denominator:
$y=\frac{\sin A+\cos A}{\sin A\cos A}$

Substitute $x=\sin A+\cos A$ in the numerator:
$y=\frac{x}{\sin A\cos A}$

Now, substitute the value of $\sin A\cos A$ from Equation 1:
$y=\frac{x}{\left(\frac{x^2-1}{2}\right)}$
$y=\frac{2x}{x^2-1}$

Cross-multiplying to simplify:
$y\left(x^2-1\right)=2x$
$x^{2}y-y=2x$

Rearranging the terms:
$x^{2}y=y+2x$

Thus, the relationship is indeed:
$y+2x=x^{2}y$